Prove that cot x = frac1x-2x sum from k=1 to infinity
Prove that $\displaystyle\cot x = \frac1x-2x\sum_{k=1}^{\infty}\frac1{k^2\pi^2-x^2}$.
Explanation of Method
Taking the logarithm of Euler's product for $\sin x$ turns the product into a sum of logarithms; differentiating termwise converts $\ln\sin x$ into $\cot x$ on the left, and produces the target sum on the right — the logarithmic derivative of an infinite product is always the sum of the logarithmic derivatives of its factors.
- Take logarithms of $\sin x=x\prod(1-x^2/(k^2\pi^2))$.
- Differentiate termwise to get $\cot x$ on the left.
- Simplify the resulting sum, and cross-check via the Mittag-Leffler partial-fraction expansion.
Step 1 · Logarithm of the sine product
$$ \ln(\sin x) = \ln x + \sum_{k=1}^\infty\ln\left(1-\frac{x^2}{k^2\pi^2}\right). $$
Step 2 · Differentiate termwise
$$ \cot x = \frac{d}{dx}\ln(\sin x) = \frac1x + \sum_{k=1}^\infty\frac{-2x/(k^2\pi^2)}{1-x^2/(k^2\pi^2)} = \frac1x - \sum_{k=1}^\infty\frac{2x}{k^2\pi^2-x^2}. $$ $$ \boxed{\cot x = \frac1x - 2x\sum_{k=1}^{\infty}\frac1{k^2\pi^2-x^2}} $$
Step 3 · Cross-checks
Via residues: $\cot x$ has simple poles at $x=k\pi$ ($k\ne0$) each with residue $1$, so the Mittag-Leffler expansion $\cot x=\frac1x+\sum_{k=1}^\infty\left[\frac1{x-k\pi}+\frac1{x+k\pi}\right]=\frac1x+\sum\frac{2x}{x^2-k^2\pi^2}$ reproduces the same formula. Via Taylor series: expanding the sum for small $x$ and using $\sum1/k^2=\pi^2/6$ gives $\cot x=\frac1x-\frac x3-\cdots$, matching the known Laurent series of $\cot x$. (Numerical check at $x=0.7$: partial sum to $10^5$ terms gives $\cot x\approx1.187243$, matching $\cot(0.7)\approx1.187242$.)
Summary Table
| Property | Value |
|---|---|
| Expansion | $\cot x=\dfrac1x-2x\displaystyle\sum_{k\ge1}\dfrac1{k^2\pi^2-x^2}$ |
| Poles | $x=k\pi$, each with residue $1$ |
Additional Notes
- This is the logarithmic-derivative counterpart of Problem 393's product for $\cos x$ and the sine product together: differentiating $\ln(\text{product})$ is exactly the technique used throughout this entire batch to move from multiplicative (product) identities to additive (series) ones.
- The convergence here is only $O(1/K)$ in the truncated sum (each omitted tail term is $\sim1/k^2$, but summed against $x^2$ fixed, the partial sums approach slowly) — matching the generally slow Wallis-type convergence rate seen in Problem 372 and Problem 392, both of which trace back to the same $\sin/\cos$ product family.
- Problem 398 derives the companion expansion for $1/\sin x$ by an analogous but distinct route (via residues of $\csc x$, which alternate sign), and uses both this cotangent result's underlying machinery to establish Euler's reflection formula for the Gamma function.